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Chapter 23: Problem 47

An unknown charge sits on a conducting solid sphere of radius \(10 \mathrm{~cm}\). If the electric field \(15 \mathrm{~cm}\) from the center of the sphere has the magnitude \(3.0 \times 10^{3} \mathrm{~N} / \mathrm{C}\) and is directed radially inward, what is the net charge on the sphere?

Short Answer

Expert verified
The net charge on the sphere is approximately \(-7.49 \times 10^{-12} \mathrm{~C}\).

Step by step solution

01

Identify the Formula

To find the net charge on the sphere, we use Gauss's Law which states that the electric field times the area over a Gaussian surface equals the charge enclosed divided by the permittivity constant. The formula is: \[ E \cdot A = \frac{Q}{\varepsilon_0} \] where \(E\) is the electric field, \(A\) is the surface area of a sphere, \(Q\) is the charge, and \(\varepsilon_0 = 8.85 \times 10^{-12} \mathrm{~C}^2/\mathrm{N} \cdot \mathrm{m}^2\) is the permittivity of free space.
02

Calculate the Surface Area

The electric field is given at \(15 \mathrm{~cm}\). We will use this as the radius of our Gaussian surface. The surface area \(A\) of a sphere is given by the formula \( A = 4\pi r^2 \). Substituting the radius, \( r = 0.15 \mathrm{~m}\): \[ A = 4\pi (0.15)^2 = 0.02827 \mathrm{~m}^2 \].
03

Insert Known Values into Gauss's Law

Insert the values into Gauss's Law formula to solve for the charge \(Q\). We have \(E = 3.0 \times 10^3 \mathrm{~N}/\mathrm{C}\), \(A = 0.02827 \mathrm{~m}^2\), and \(\varepsilon_0 = 8.85 \times 10^{-12} \mathrm{~C}^2/\mathrm{N} \cdot \mathrm{m}^2\). Thus, \[ Q = E \cdot A \cdot \varepsilon_0 = (3.0 \times 10^3) \cdot (0.02827) \cdot (8.85 \times 10^{-12}) \].
04

Calculate the Net Charge

Compute \( Q \) using the values from the previous step.\[ Q = (3.0 \times 10^3) \times 0.02827 \times 8.85 \times 10^{-12} \approx 7.49 \times 10^{-12} \mathrm{~C} \]. The charge is negative since the electric field is directed radially inward.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

electric field
The electric field is a vector field that surrounds electrically charged particles and exerts force on other charges in the field. Imagine it as an invisible force field around a charged object.
If you place another charged object within this field, it will experience an attractive or repulsive force. The direction of the electric field is the direction of the force that a positive test charge would feel.
In this particular exercise, the electric field is indicated as radial and directed inward, suggesting that the test charge would move towards the sphere. The magnitude of the electric field in our example is given as \(3.0 \times 10^{3} \text{ N/C}\). This tells us how strong the force will be felt per unit charge.
  • Electric fields can be uniform or vary in space.
  • Fields can be calculated using Gauss's Law, especially in symmetrical setups like spheres and cylinders.
permittivity of free space
Permittivity of free space is a crucial constant in electromagnetism, denoted by \(\varepsilon_0\). It quantifies how much electric field is "permitted" to penetrate through free space, influencing how electric charges interact. Mathematically, it appears in equations that describe electric fields, like Gauss's Law.
  • Its value is \( 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2\).
  • It acts as a proportionality constant that relates the electric field and electric flux to the charge enclosed by a Gaussian surface.
Understanding \(\varepsilon_0\) helps calculate how easily an electric field can influence a charge in space, significantly impacting the computation of forces in electric fields. In this exercise, \(\varepsilon_0\) is vital for finding the net charge via Gauss鈥檚 Law.
net charge calculation
Net charge calculation involves determining the total charge on an object based on electric field measurements. Using Gauss's Law is the most effective way in scenarios with high symmetry, such as spheres. Gauss's Law states:\[ E \cdot A = \frac{Q}{\varepsilon_0} \]Here, \(E\) is the electric field, \(A\) is the surface area of the Gaussian surface, \(Q\) is the net charge, and \(\varepsilon_0\) is the permittivity of free space.
In our case, the goal is to solve for \(Q\), the charge, knowing \(E\), \(A\), and \(\varepsilon_0\).
The steps are:
  • Calculate the surface area \(A\) using the radius of the Gaussian surface, which in this problem is \(0.15 \text{ m}\).
  • Insert the known values into the Gauss's Law equation.
  • Perform the computation to find the charge \(Q\).
This process ultimately gives us the charge on the conducting sphere, in this case, \(-7.49 \times 10^{-12} \text{ C}\). The negative sign indicates the direction of the electric field is inward, signifying a negatively charged sphere.

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Most popular questions from this chapter

Charge \(Q\) is uniformly distributed in a sphere of radius \(R\). (a) What fraction of the charge is contained within the radius \(r=R / 2.00 ?\) (b) What is the ratio of the electric field magnitude at \(r=R / 2.00\) to that on the surface of the sphere?

Explosions ignited by electrostatic discharges (sparks) constitute a serious danger in facilities handling grain or powder. Such an explosion occurred in chocolate crumb powder at a biscuit factory in the 1970 s. Workers usually emptied newly delivered sacks of the powder into a loading bin, from which it was blown through electrically grounded plastic pipes to a silo for storage. Somewhere along this route, two conditions for an explosion were met: (1) The magnitude of an electric field became \(3.0 \times 10^{6} \mathrm{~N} / \mathrm{C}\) or greater, so that electrical breakdown and thus sparking could occur. ( 2 ) The energy of a spark was \(150 \mathrm{~mJ}\) or greater so that it could ignite the powder explosively. Let us check for the first condition in the powder flow through the plastic pipes. Suppose a stream of negatively charged powder was blown through a cylindrical pipe of radius \(R=5.0 \mathrm{~cm}\). Assume that the powder and its charge were spread uniformly through the pipe with a volume charge density \(\rho\). (a) Using Gauss' law, find an expression for the magnitude of the electric field \(\vec{E}\) in the pipe as a function of radial distance \(r\) from the pipe center. (b) Does \(E\) increase or decrease with increasing \(r ?\) (c) Is \(\vec{E}\) directed radially inward or outward? (d) For \(\rho=1.1 \times 10^{-3} \mathrm{C} / \mathrm{m}^{3}\) (a typical value at the factory), find the maximum \(E\) and determine where that maximum field occurs. (e) Could sparking occur, and if so, where? (The story continues with Problem 70 in Chapter \(24 .\) )

A spherical conducting shell has a charge of \(-14 \mu \mathrm{C}\) on its outer surface and a charged particle in its hollow. If the net charge on the shell is \(-10 \mu \mathrm{C},\) what is the charge (a) on the inner surface of the shell and (b) of the particle?

A long, nonconducting, solid cylinder of radius \(4.0 \mathrm{~cm}\) has a nonuniform volume charge density \(\rho\) that is a function of radial distance \(r\) from the cylinder axis: \(\rho=A r^{2} .\) For \(A=2.5 \mu \mathrm{C} / \mathrm{m}^{5},\) what is the magnitude of the electric field at (a) \(r=3.0 \mathrm{~cm}\) and (b) \(r=5.0 \mathrm{~cm} ?\)

Charge of uniform volume density \(\rho=1.2 \mathrm{nC} / \mathrm{m}^{3}\) fills an infinite slab between \(x=-5.0 \mathrm{~cm}\) and \(x=+5.0 \mathrm{~cm} .\) What is the magnitude of the electric field at any point with the coordinate (a) \(x=4.0 \mathrm{~cm}\) and (b) \(x=6.0 \mathrm{~cm} ?\)
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